Gibbs Free Energy and Reaction Equilibrium Practice Problems (1/2)
Key Concepts in Gibbs Free Energy Calculations
Gibbs Free Energy (∆G)
Definition: Gibbs Free Energy is a thermodynamic potential that measures the maximum reversible work that can be performed by a thermodynamic system at constant temperature and pressure.
Formula: ∆G = ∆H - T∆S Where:
∆H = Change in enthalpy
T = Temperature in Kelvin
∆S = Change in entropy
Standard Gibbs Free Energy of Formation (AG°f)
Definition: The change in Gibbs Free Energy when one mole of a compound is formed from its elements in their standard states.
Values Provided:
2 HNO3(aq) : -110.9 kJ/mol
NO(g) : 87.6 kJ/mol
NO2(g) : 51.3 kJ/mol
H2O(l) : -237.1 kJ/mol
Reaction Equilibrium Constant (K)
Formula: K = exp(-∆G°/RT) Where:
R = Universal gas constant (8.314 J/(mol·K))
T = Temperature in Kelvin
Example Calculation for AG
Given Reaction:
2 Hg(g) + O2(g) → 2 HgO(s)Provided Data:
AG = -180.8 kJ
P(Hg) = 0.025 atm
P(O2) = 0.037 atm
Utilize Gibb's Energy and Equilibrium:
Calculate standard Gibbs free energy for the formation of products and reactants and apply to find K.
Example Calculation for ∆G at Non-standard Conditions
Given Reaction:
N2(g) + 3 H2(g) ⇌ 2 NH3(g)Conditions:
PN2 = 33.0 atm
PH2 = 99.0 atm
PNH3 = 2.0 atm
Calculate ∆G at 427°C (700 K):
Apply the formula: ∆G = ∆G° + RT ln(Q) Where:
Q = (P(NH3)²)/(P(N2) * P(H2)³)
Steps for Solving Problems
Identify Given Data: Note all necessary variables and values given.
Identify Reaction Components: Clearly establish reactants and products.
Choose the Correct Formula: Based on the problem situation (Equilibrium, Standard, or Non-standard conditions).
Perform Calculations: Execute the mathematical operations step by step, ensuring units are consistent.
Interpret Results: Understanding the implications of the value of ∆G (negative, positive, or zero) as it relates to spontaneity and equilibrium.